Question

Find all the zeroes of 2x^4- 9x^3 + 5x^2 +3x -1 if two of its zeroes are 2+3root and 2-3root.

Answer 1

Heya !!

Here's your answer.. ⬇⬇
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p( x ) = 2x^4 - 9x^3 + 5x^2 + 3x - 1

➡ Given that zeroes are 2 + √3 and 2 - √3.

x = 2 + √3 and x = 2 - √3

x - 2 - √3 = 0 and x - 2 + √3 = 0

( x - 2 ) - √3 = 0 and ( x - 2 ) + √3 = 0

= { ( x - 2 ) - √3 }{ ( x - 2 ) + √3 }

= ( x - 2 )^2 - ( √3 )^2
【 a^2 - b^2 = ( a-b )( a+b )】

= x^2 - 4x + 4 - 3

= x^2 - 4x + 1

➡ Now, dividing p( x ) by x^2 - 4x + 1 we get..,
[ Division is in the attachment ]

Factorize the polynomial..
2x^2 - x - 1 = 0
2x^2 - 2x + x - 1 = 0
2x ( x - 1 ) + 1 ( x - 1 ) = 0
( 2x + 1 ) ( x - 1 ) = 0
( 2x + 1 ) = 0 and ( x - 1 ) = 0
x = -1/2 and x = 1

The other remaining zeroes are -1/2 and 1.
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Hope it helps

Thanks :))

Answer 2

Answer:

Hope this helps you ✌️✌️